Question

giving a medals!please help!

which of the folowing must be true for all values of x?
I.(X+1)^2 is greater than or equal to x^2
II.(x-2)^2 is greater than or equal to 0
III.x^2 +1 is greater than or equal to 2x

Answer 1

@jim_thompson5910 and @CrazyMad plz help

Answer 2

are you able to graph each expression?

Answer 3

this is a question on the psat so i don't think i will be able to use a graphing calculator.

Answer 4

what does (x+1)^2 expand to?

Answer 5

x^2 +2x +1,right?

Answer 6

it is the 2nd option as anything in side the ( ) will be turned positive once it is squared

Answer 7

correct

Answer 8

so

\[\Large (x+1)^2 \ge x^2\]

turns into

\[\Large x^2+2x+1 \ge x^2\]

Answer 9

mhmm.. i'm starting to get it

Answer 10

we can then subtract x^2 from both sides to get \[\Large 2x+1 \ge 0\]

Answer 11

what do you get once you solve for x?

Answer 12

x> or equal to -1/2

Answer 13

so \[\Large (x+1)^2 \ge x^2\] is only true if \[\Large x \ge -\frac{1}{2}\]

Answer 14

therefore, inequality I is not true for all real numbers

Answer 15

yes...

Answer 16

ok...so do we do the same thing for all 3?

Answer 17

ii) Whenever you square a real number or a real function it is always greater than or equal to zero: { f(x) }^2 >= 0
Therefore \((x-2)^2 \ge 0\) (always).

Answer 18

agreed with aum on #2

squaring any real number will never result in a negative number

Answer 19

iii)\[
(x-1)^2 \ge 0 \\
x^2 - 2x + 1 \ge 0 \\
x^2 + 1 \ge 2x
\]

Answer 20

so how do i know which one is true for all real numbers?

Answer 21

ii) and iii)

Answer 22

and can u explain why 3 isn't correct,again?

Answer 23

\[\Large x^2 +1 \ge 2x\]

\[\Large x^2 + 1 - 2x\ge 0\]

\[\Large x^2 -2x + 1\ge 0\]

\[\Large (x-1)^2 \ge 0\]

Answer 24

(x-1)^2 is never negative, so it is either 0 or greater than 0

Answer 25

x2+1≥2x


x2+1−2x≥0


x2−2x+1≥0


(x−1)2≥0

Answer 26

oops...sorry,lol.

Answer 27

yeah it looks like you got it

Answer 28

haha:)
but, (x-1)^2 isn't even one of the options.

Answer 29

it looks like (x−1)2≥0 is trying to say \[\Large (x-1)^2 \ge 0\]

Answer 30

refresh the page if you get any crazy looking symbols

Answer 31

the thing is,(x-1)^2 isn't one of the 3 options i gave u,so where did it come from?

Answer 32

what are the options again?

Answer 33

I.(X+1)^2 is greater than or equal to x^2
II.(x-2)^2 is greater than or equal to 0
III.x^2 +1 is greater than or equal to 2x

Answer 34

I got \[\Large (x-1)^2 \ge 0\] starting from \[\Large x^2 +1 \ge 2x\]

Answer 35

I show the steps above

Answer 36

We have to prove or disprove x^2 +1 is greater than or equal to 2x
x^2 + 1 >= 2x
x^2 + 1 - 2x >= 0
(x-1)^2 >= 0

Any function square will always be greater than equal to zero for ALL real values of x.
Therefore, (x-1)^2 >= 0 is always true which makes x^2 + 1 >= 2x because we derived (x-1)^2 >=0 from x^2 + 1 >= 2x

Answer 37

you get (x-1)^2 from factoring x^2-2x+1

Answer 38

ok guyz...so 3 is always true,right?

Answer 39

Yes, 2 and 3 are always true.

Answer 40

but 1 is false because it has to be >= 0,right?

Answer 41

it's only true when x >= -1/2

Answer 42

otherwise, it's false

Answer 43

so overall, it's false for all real numbers

Answer 44

so the final answer is:both 2 and 3 are true.

Answer 45

@jim_thompson5910 ??

Answer 46

yes only 2 and 3 are true

Answer 47

thanku sooo much guyz..
what level of math r u in?